It's better to deal with equations than inequalities. Stick with [tex]1 + (a_2 + d_2) + (a_2d_2 - c_2b_2) = 0.[/tex] Start setting variables to zero until only two unknowns remain and see what happens. You do have the constraints that [itex]a_2d_2[/itex] and [itex]b_2c_2[/itex] are not both zero and none of the variables can be negative.

So I can manageably get to the equation
[itex]1+(a_2+d_2)+(a_2d_2-c_2b_2)=0[/itex]
by setting A as the identity Matrix.

I see that there is a connection in the sense that:
[itex](a_2d_2-c_2b_2) =/= 0[/itex]

So we should be able to represent it as an unknown variable, say, x.

[itex] 1+(a_2+d_2)+x=0.[/itex]

I would write [tex]1 + (a_2 + d_2 + a_2d_2) = (b_2c_2).[/tex] The left hand side is at least 1. You can, by appropriate choice of [itex]a_2\geq 0[/itex] and [itex]d_2 \geq 0[/itex], set it to be exactly 1.